12 - Sum of the Arithmetic Series (Algebra)

In this post, you can check the solutions to the exercises of Sum of the Arithmetic Series of Algebra chapter of Readmore Optional Mathematics by DR Simkhada. It consists of complete answers to all odd questions mentioned in the textbook.

Before starting, make sure to check the following note of sum of the Arithmetic Series chapter.

Things to remember:

  1. The first term of the sequence is denoted by $t_1$ or a.
  2. The common difference of the sequence is denoted by d.
  3. The last term of the sequence is denoted by b or l.
  4. The formula for sum of arithmetic series is $S_n = \frac{n}{2} (a + b)$
  5. Another formula is $S_n = \frac{n}{2} [2a + (n-1)d].



1. Find the sum of the following series:

a) 5+11+17+.... to 20 terms.
Solution:

Here,
a = 5
d = 11-5 = 6
n = 20

Using formula,
$S_n = \frac{n}{2} [2a + (n-1)d]$

$= \frac{20}{2} [2×5 + (20-1)6]$

$= 10 × [10 + 19×6]$

$= 10 × 124$

$= 1240$

Hence, the required sum of the series is 1240.


c) -3 +1 +5 +...... to 50 terms
Solution:

Here,
a = -3
d = 1 - (-3) = 4
n = 50

Using formula,
$S_n = \frac{n}{2} [2a + (n-1)d]$

$= \frac{50}{2} [2(-3) + (50-1)4]$

$= 25 [ -6 + 49×4]$

$= 25 × [-6 + 196]$

$= 25 × 190$

$= 4750$


7 a) Find the common difference and first term of an arithmetic series if the sum of the first 6 terms and 9 terms are 183 and 369 respectively.
Solution:

Here,
$S_6 = 183$
$S_9 = 369$

We know,
$S_n = \frac{n}{2} [2a + (n-1)d]$

Now,
$S_6 = \frac{6}{2} [2a + (6-1)d]$

$or, 183 =  3 × [2a + 5d]

$or, \frac{183}{3} = 2a + 5d$

$or, 61 = 2a + 5d$

$or, a = \frac{61 - 5d}{2}$ ----- (i)

And,
$S_9 = \frac{9}{2} [2a + (9-1)d]$

$or, 369 = \frac{9}{2} [2a + 8d]$

$or, \frac{369}{9} = \frac{1}{2} × 2[a +4d]$

$or, 41 = a + 4d$

$or, a = 41 - 4d$ ----- (ii)

From equations (i) and (ii), we get,
$or, \frac{61 - 5d}{2} = 41 - 4d$

$or, 61 - 5d = 2(41-4d)$

$or, 61 - 5d = 82 - 8d$

$or, 8d - 5d = 82 - 61$

$or, 3d = 21$

$\therefore d = 7$


Put value of d in equation (ii), we get,
$or, a = 41 - 4×7$

$or, a = 41 - 28$

$\therefore a = 13$

Hence, the required common difference and the first term of the arithmetic series are 7 and 13, respectively.


9 a) The sum of three numbers in an AS is 39 and their product is 2184. Find the numbers.
Solution:

Let the three numbers in an AS be (a -d), a, (a+d).

According to the question,
$(a -d) + a + (a -d) = 39$

$or, 3a = 39$

$\therefore a =13$


Also,
$(a -d)a(a+d) = 2184$

$or, a (a^2 - d^2) = 2184$

$or, 13(13^2 - d^2) = 2184$

$or, (169 - d^2) = \frac{2184}{13}$

$or, 169 - d^2 = 168$

$or, d^2 = 1$

$\therefore d =\pm 1$


When d = 1,
(a - d) = (13 -1) = 12
a = 13
(a + d) = (13+1) = 14

When d = -1,
(a - d) = (13 +1) = 14
a = 13
(a + d) = (13 -1) = 13

Hence, the possible numbers are 12,13,14 or 14,13,12.

About Readmore Optional Mathematics Book 10

Author: D. R. Simkhada
Editor: I. R. Simkhada

Readmore Publishers and Distributors
T.U. Road, Kuleshwor - 14, Kathmandu, Nepal
Phone: 4672071, 5187211, 5187226
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Sum of the Arithmetic Series - Class 10 Solved Exercises | Readmore Optional Mathematics is a collection of the solutions related to exercises of Sum of the Arithmetic Series chapter from Algebra Unit for Nepal's Secondary Education Examination (SEE) appearing students.

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